Equivalence of definitions

The set of all linear combinations of a subset S of V, a vector space over K, is the smallest linear subspace of V containing S.

Proof. We first prove that span S is a subspace of V. Since S is a subset of V, we only need to prove the existence of a zero vector 0 in span S, that span S is closed under addition, and that span S is closed under scalar multiplication. Letting ${\displaystyle S=\{\mathbf {v} _{1},\mathbf {v} _{2},\ldots ,\mathbf {v} _{n}\}}$, it is trivial that the zero vector of V exists in span S, since ${\displaystyle \mathbf {0} =0\mathbf {v} _{1}+0\mathbf {v} _{2}+\cdots +0\mathbf {v} _{n}}$. Adding together two linear combinations of S also produces a linear combination of S: ${\displaystyle (\lambda _{1}\mathbf {v} _{1}+\cdots +\lambda _{n}\mathbf {v} _{n})+(\mu _{1}\mathbf {v} _{1}+\cdots +\mu _{n}\mathbf {v} _{n})=(\lambda _{1}+\mu _{1})\mathbf {v} _{1}+\cdots +(\lambda _{n}+\mu _{n})\mathbf {v} _{n}}$, where all ${\displaystyle \lambda _{i},\mu _{i}\in K}$, and multiplying a linear combination of S by a scalar ${\displaystyle c\in K}$ will produce another linear combination of S: ${\displaystyle c(\lambda _{1}\mathbf {v} _{1}+\cdots +\lambda _{n}\mathbf {v} _{n})=c\lambda _{1}\mathbf {v} _{1}+\cdots +c\lambda _{n}\mathbf {v} _{n}}$. Thus span S is a subspace of V.

Suppose that W is a linear subspace of V containing S. It follows that ${\displaystyle S\subseteq \operatorname {span} S}$, since every v_i is a linear combination of S (trivially). Since W is closed under addition and scalar multiplication, then every linear combination ${\displaystyle \lambda _{1}\mathbf {v} _{1}+\cdots +\lambda _{n}\mathbf {v} _{n}}$ must be contained in W. Thus, span S is contained in every subspace of V containing S, and the intersection of all such subspaces, or the smallest such subspace, is equal to the set of all linear combinations of S.

Size of spanning set is at least size of linearly independent set

Every spanning set S of a vector space V must contain at least as many elements as any linearly independent set of vectors from V.

Proof. Let ${\displaystyle S=\{\mathbf {v} _{1},\ldots ,\mathbf {v} _{m}\}}$ be a spanning set and ${\displaystyle W=\{\mathbf {w} _{1},\ldots ,\mathbf {w} _{n}\}}$ be a linearly independent set of vectors from V. We want to show that ${\displaystyle m\geq n}$.

Since S spans V, then ${\displaystyle S\cup \{\mathbf {w} _{1}\}}$ must also span V, and ${\displaystyle \mathbf {w} _{1}}$ must be a linear combination of S. Thus ${\displaystyle S\cup \{\mathbf {w} _{1}\}}$ is linearly dependent, and we can remove one vector from S that is a linear combination of the other elements. This vector cannot be any of the w_i, since W is linearly independent. The resulting set is ${\displaystyle \{\mathbf {w} _{1},\mathbf {v} _{1},\ldots ,\mathbf {v} _{i-1},\mathbf {v} _{i+1},\ldots ,\mathbf {v} _{m}\}}$, which is a spanning set of V. We repeat this step n times, where the resulting set after the pth step is the union of ${\displaystyle \{\mathbf {w} _{1},\ldots ,\mathbf {w} _{p}\}}$ and m - p vectors of S.

It is ensured until the nth step that there will always be some v_i to remove out of S for every adjoint of v, and thus there are at least as many v_i's as there are w_i's—i.e. ${\displaystyle m\geq n}$. To verify this, we assume by way of contradiction that ${\displaystyle m<n}$. Then, at the mth step, we have the set ${\displaystyle \{\mathbf {w} _{1},\ldots ,\mathbf {w} _{m}\}}$ and we can adjoin another vector ${\displaystyle \mathbf {w} _{m+1}}$. But, since ${\displaystyle \{\mathbf {w} _{1},\ldots ,\mathbf {w} _{m}\}}$ is a spanning set of V, ${\displaystyle \mathbf {w} _{m+1}}$ is a linear combination of ${\displaystyle \{\mathbf {w} _{1},\ldots ,\mathbf {w} _{m}\}}$. This is a contradiction, since W is linearly independent.

Spanning set can be reduced to a basis

Let V be a finite-dimensional vector space. Any set of vectors that spans V can be reduced to a basis for V, by discarding vectors if necessary (i.e. if there are linearly dependent vectors in the set). If the axiom of choice holds, this is true without the assumption that V has finite dimension. This also indicates that a basis is a minimal spanning set when V is finite-dimensional.

Notes

The linear span of a set is dense in the closed linear span. Moreover, as stated in the lemma below, the closed linear span is indeed the closure of the linear span.

Closed linear spans are important when dealing with closed linear subspaces (which are themselves highly important, see Riesz's lemma).

A useful lemma

Let X be a normed space and let E be any non-empty subset of X. Then

(So the usual way to find the closed linear span is to find the linear span first, and then the closure of that linear span.)

Textbooks

• Axler, Sheldon Jay (2015). Linear Algebra Done Right (3rd ed.). Springer. ISBN 978-3-319-11079-0.
• Hefferon, Jim (2020). Linear Algebra (4th ed.). Orthogonal Publishing. ISBN 978-1-944325-11-4.
• Lane, Saunders Mac; Birkhoff, Garrett (1999) [1988]. Algebra (3rd ed.). AMS Chelsea Publishing. ISBN 978-0821816462.
• Roman, Steven (2005). Advanced Linear Algebra (2nd ed.). Springer. ISBN 0-387-24766-1.
• Rynne, Brian P.; Youngson, Martin A. (2008). Linear Functional Analysis. Springer. ISBN 978-1848000049.
• Lay, David C. (2021) Linear Algebra and Its Applications (6th Edition). Pearson.

Web

• Lankham, Isaiah; Nachtergaele, Bruno; Schilling, Anne (13 February 2010). "Linear Algebra - As an Introduction to Abstract Mathematics" (PDF). University of California, Davis. Retrieved 27 September 2011.
• Weisstein, Eric Wolfgang. "Vector Space Span". MathWorld. Retrieved 16 Feb 2021.
• "Linear hull". Encyclopedia of Mathematics. 5 April 2020. Retrieved 16 Feb 2021.